# Heat Equation on $[0,T] \times \mathbb{R}^n$

I'm currently looking for a complete proof of a classical result (very useful for viscosity methods) and surprisingly all the references I can get study the heat equation on bounded domain.

Do you know where I can find a complete proof of such a result : There exists a unique solution $u$ to the problem

$\left\{ \begin{array}{rcll} \partial_t u(t,x) - \eta \Delta u(t,x) &=& f(t,x)&\qquad\text{on } ]0,T]\times \mathbb{R}^n \\ u(0,x) &=& u_0(x)&\qquad\text{on } \mathbb{R}^n \end{array} \right.$

with $f\in L^2(0,T;L^2(\mathbb{R}^n))$, $u_0 \in H^1(\mathbb{R}^n)$ and $\eta>0$ fixed and furthermore $u\in L^2(0,T;H^2(\mathbb{R}^n))$ and $\partial_t u \in L^2(0,T;L^2(\mathbb{R}^n))$.

Thank you

Fourier transform in $x$ gets you there: $\dot v+\eta\vert\xi\vert^2 v=g(t,\xi),\quad v(0)=v_0,$ so that $$v(t,\xi)=e^{-t\eta \vert\xi\vert^2} v_0(\xi)+\int_0^te^{-(t-s)\eta \vert\xi\vert^2} g(s,\xi) ds.$$ Since $\iint_0^T\vert g(t,\xi)\vert^2 dtd\xi<+\infty$ and $\int\vert v_0(\xi)\vert^2 d\xi<+\infty$, you get $$\iint_0^Te^{-2t\eta \vert\xi\vert^2} \vert \xi\vert^2\vert v_0(\xi)\vert^2 dt d\xi\le \int\frac{[e^{-2t\eta \vert \xi\vert^2}]^{t=0}_{t=T}}{2\eta}\vert v_0(\xi)\vert^2 d\xi\le (2\eta)^{-1}\int\vert v_0(\xi)\vert^2 d\xi<+\infty,$$ and similarly with $H=\mathbf 1_{\mathbb R_+}$ $$\vert \xi\vert^2\int H(t-s)e^{-(t-s)\eta \vert \xi\vert^2} g(s)H(s) ds= \vert \xi\vert^2 H(t)e^{-t\eta \vert \xi\vert^2} \ast g(t)H(t)$$ with $L^2$ norm (in $t$) bounded above by $$\vert \xi\vert^2 (\int_0^T\vert g(t,\xi)\vert^2 dt)^{1/2}\int_0^{+\infty}e^{-t\eta \vert \xi\vert^2} dt =\eta^{-1}(\int_0^T\vert g(t,\xi)\vert^2 dt)^{1/2}$$ and the square of the $L^2$ norm (in $t,\xi$) is bounded above by $$\eta^{-2}\iint_0^T\vert g(t,\xi)\vert^2 dtd\xi<+\infty.$$ The last statement on $\dot v$ comes from the equation and the previous results.